$2 \ mol$ of $N_2$ is mixed with $6 \ mol$ of $H_2$ in a closed vessel of $1 \ L$ capacity. If $50\%$ of $N_2$ is converted into $NH_3$ at equilibrium,the value of $K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is

For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a $1 \ L$ closed vessel and heated until the following equilibrium is attained:
${N_2}{O_{4(g)}} \rightleftharpoons 2N{O_{2(g)}}$
If $50\%$ of $N_2O_{4(g)}$ dissociates at equilibrium,what will be the equilibrium constant (in $mol \ L^{-1}$)? (Mol. wt. of $N_2O_4 = 92$)

Given the equilibria:
$I: A + 2B \rightleftharpoons C ; K_{eq} = K_1$
$II: C + D \rightleftharpoons 3A ; K_{eq} = K_2$
$III: 6B + D \rightleftharpoons 2C ; K_{eq} = K_3$
Which of the following relations is correct?

For the reaction $P + Q \rightleftharpoons R + C$,the equilibrium constant $K_c$ is $10^{-2}$ and the forward rate constant $K_f$ is $10^{-1}$. The rate constant for the backward reaction $(K_b)$ will be:

$x A_{(s)} \rightleftharpoons y B_{(g)} + z C_{(g)}$. If $\frac{K_c}{K_p} = (RT)^{-2}$,then which is correct?